Question

How do you determine the convergence or divergence of `Sigma ((-1)^(n+1)n^2)/(n^2+5)` from `[1,oo)`?

Answer 1

The non-alternating portion of the sequence that makes up the series is `n^2/(n^2+5)`.

Note that `lim_(nrarroo)n^2/(n^2+5)=1`.

This means that as `n` grows infinitely larger, the series will begin to resemble `sum_(n=1)^oo(-1)^n`, which diverges because it never "settles".

A powerful tool for testing divergence is the rule that if `suma_n` converges, then `lim_(nrarroo)a_n=0`. That is, as `n` gets larger, the terms of any convergent series will approach `0`.

Here, `lim_(nrarroo)a_n=1`, so the series is divergent.